Length of one diagonal of a rhombus is $$24\,\text{cm}$$. The lengths of the other diagonal and its side are both integers. How many such rhombus are possible?

The diagonals bisect each other at right angles, so if the other diagonal is $$d$$ and the side is $$s$$, then $$s^2 = 12^2 + \left(\frac{d}{2}\right)^2$$, that is $$(2s - d)(2s + d) = 576$$. Writing $$2s - d = 2p$$ and $$2s + d = 2q$$ gives $$pq = 144$$ with $$p + q$$ even, and the pairs $$(2, 72)$$, $$(4, 36)$$, $$(6, 24)$$, $$(8, 18)$$ work. These give $$d = 70, 32, 18, 10$$ with $$s = 37, 20, 15, 13$$, so 4 rhombuses are possible.